Weakly-Initial Set of Covers

Definition

A cover of a set $Y$ is just a surjective map from another set $X$:

$$\mathrm{Cov}(Y)=\sum _{X:U}X\twoheadrightarrow Y$$ Link to original

Definition

The category of covers of $Y$ is given as:

• Objects are covers in $\mathrm{Cov}(Y)$
• $\mathrm{Hom}[(X,p),(Z,q)]={\sum }_{f:X\to Z}q\circ f=p$

Link to original

Definition

A weakly-initial set in an arbitrary category with families $\mathcal{C}$ is a family $(I,\bar{X})$ such that:

$$\forall Z:\mathcal{C}.\exists i\in I.\exists f:X_i\to Z$$Link to original

Definition

A weakly initial set of covers over $Y$ is just a weakly initial set in the category of covers $\mathbf{Cov}(Y)$.

That is:

$$\begin{array}{rl}& \text{Suppose }(I,\bar{X}):\prod _{i:I}\mathrm{Cov}(Y).\\ & \text{Say }(I,\bar{X})\text{ is WISC for Y iff}\\ & \forall q:Z\twoheadrightarrow Y\\ & \exists i:I;f:X_i\to Z\\ & \text{s.t. }q\circ f\text{ is a surjection}\end{array}$$

Remarks

Full choice, such as ZFC implies that identities are a weakly-initial object of $\mathrm{Cov}(Y)$. Choice tells that every fibre is contractible. I.e. the surjection has a section. Without choice, this object might not be weakly initial since the section amounts to a unique choice.

The fact that it’s a surjection is important because otherwise there is a base-point with an empty fibre. Meaning there’s no choice function.

Independence from ZF

Since AC implies WISC, then the consistency of ZFC implies ZF is consistent with WISC. It has been proven that WISC isn’t entailed by ZF. berg2012-wisc The proof works by:

1. Given any polynomial family of $\mathbb{N}$ covers $B:I\to \mathcal{U}$
2. Extend it to by adding, 0 and 1.
3. The let $W$ be the W type of $A◁B$
4. Define $m:W\to \mathrm{Ord}$ by transfinite induction s.t.

\begin{aligned} m(sup0_) &= 0\ m(sup1t) &= 1+t(x)\ m(supit) &= \sup {m(tb).b:B i} \end{aligned}